Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

An elementary problem asks for Arg($f(z)$) after a single complete counter-clockwise rotation of the point z about the origin, beginning at the point z = 2 and taking the angle there to be 0, with $f(z) = \sqrt{z^2+2z-3}.$

Plotting a few points by hand and then by computer, I find the final value as Arg(f(z)) = 0. The answer in the text is $\pi.$

I am probably overlooking something obvious...thanks for any insight.

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

Think about $g(z) = \sqrt{z}$. After you go through a complete counter-clockwise rotation, you find that

$$g(z e^{i 2 \pi}) = g(z) e^{i \pi} $$

That is, a $2 \pi$ rotation in $z$ produces a $\pi$ rotation in $\sqrt{z}$. Think about this in light of your function.

share|improve this answer
    
Try it without the linear term and see what you get. –  Ron Gordon Jan 10 '13 at 22:14
    
Yes, so you may conclude that the linear term introduces the $\pi$ phase. This is how a Riemann surface works, by the way. –  Ron Gordon Jan 10 '13 at 22:21
    
Sorry, I was just being suggestive. Maybe I am misreading your last comment, but $z^2+2 z-3 < 0$ when $z \in (-3,1)$. There is a branch cut (another way of looking at the complex plane) in the plane that is this interval. For $|z| = 2$, you should see the $\pi$ phase after a $2 \pi$ rotation. But, if you do not breach the branch, you will not see such a rotation. –  Ron Gordon Jan 10 '13 at 22:36
    
It is $\pi$ because it crosses the branch cut between $z \in (-3,1)$. –  Ron Gordon Jan 10 '13 at 22:44
    
See this: en.wikipedia.org/wiki/Branch_cut#Branch_cuts –  Ron Gordon Jan 10 '13 at 22:46
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.